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In an experiment on photoelectric effect light of wavelength $400 n m$ is incident on a caesium plate at the rate of $5 W$ . The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every $10^{6}$ photons is able to eject a photoelectron, find the photocurrent in the circuit.

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$2.6 μ A$

$6 μ A$

$1.6 μ A$

$4 μ A$

answer is D.

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Detailed Solution

$E = \frac{12375}{4000} = 3.1 e V$

Number of photoelectrons emitted per second

$n = (\frac{1}{10^{6}}) (\frac{5}{3.1 \times 1.6 \times 10^{- 19}}) = 1 \times 10^{13} persecond$

Since,

$i = \frac{q}{t} = \frac{N e}{t} = (\frac{N}{t}) e = n e$

$\Rightarrow i = (n e) = 1 \times 10^{13} \times 1.6 \times 10^{- 19}$

$\Rightarrow i = 1.6 \times 10^{- 6} A = 1.6 μ A$

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